For the expansion (1+x)^(n) find the sum of the products of the binomial coefficients taken two at a time?

2017-02-15 1:42 am

回答 (2)

2017-02-15 2:27 am
✔ 最佳答案
(1+x)²ⁿ = (1+x)ⁿ * (x+1)ⁿ = ∑(r=0,n) C(n,r)x ͬ * ∑(k=0,n) C(n,k)xⁿ⁻ᵏ … (i)

Set x= 1 to get

2²ⁿ = ∑(r=0,n) C(n,r) * ∑(k=0,n) C(n,k)

Expand RHS for 2²ⁿ = ∑(r=0,n)C(n,r)² + 2∑(r=0,n), k=0,n) (r≠k) C(n,r)C(n.k) … (ii)

The last term (2S) is what you want

To find ∑(r=0,n)C(n,r)² equate coefficients of xⁿ in (i), selecting combinations where r=k

C(2n,n) = ∑(r=k) C(n,r) C(n,k) = ∑(r=0,n) C(n,r)²

From (ii) : 2²ⁿ = C(2n,n) + 2S ⟹ S = ½( 2²ⁿ−C(2n,n) )
2017-02-15 1:52 am
The coefficients will be C(n,0), C(n,1), C(n,2), ..., C(n,n)

Answer:
C(n,0) * C(n,1) + C(n,2) * C(n,3) + ... C(n,n)

For example:
n = 1
(1 * 1) = 1

n = 3
(1 * 3) + (3 * 1) = 6

n = 5
(1 * 5) + (10 * 10) + (5 * 1) = 110

n = 7
(1 * 7) + (21 * 35) + (35 * 21) + (7 * 1) = 1484

It's not clear what you are supposed to do if you have an even number of coefficients.


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